Solveeit Logo
Login

Question

Mathematics Question on Arithmetic Progression

If sin⁡A,sin⁡B,cos⁡A\sin A , \sin B , \cos A are in GP, then roots of x2+2xcot⁡B+1=0x ^{2}+2 x \cot B +1=0 are always

A

real

B

imaginary

C

greater than 1

D

equal

Answer

real

Explanation

Solution

sin⁡A,sin⁡B,cos⁡A\sin A, \sin B, \cos A are in G.P. ∴sin⁡2B=sin⁡Acos⁡A→1\therefore \sin ^{2} B =\sin A \cos A \rightarrow 1 x2+2xcot⁡B+1=0x^{2}+2 x \cot B+1=0 D=(2cot⁡B)2−4(1)(1)D =(2 \cot B )^{2}-4(1)(1) D=4cot⁡2B−4D =4 \cot ^{2} B -4 D=4[cos⁡2Bsin⁡2B−1]D=4\left[\frac{\cos ^{2} B}{\sin ^{2} B}-1\right] D=4[cos⁡2B−sin⁡2Bsin⁡2B]D=4\left[\frac{\cos ^{2} B-\sin ^{2} B}{\sin ^{2} B}\right] D=4[1−sin⁡2B−sin⁡2Bsin⁡2B][∵cos⁡2θ+sin⁡2θ=1]D=4\left[\frac{1-\sin ^{2} B-\sin ^{2} B}{\sin ^{2} B}\right] \quad\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right] D=4[1−2sin⁡2Bsin⁡2B]D=4\left[\frac{1-2 \sin ^{2} B}{\sin ^{2} B}\right] D=4[1−2sin⁡Acos⁡Asin⁡2B]D=4\left[\frac{1-2 \sin A \cos A}{\sin ^{2} B}\right] (From equation 1) D=4[sin⁡2A+cos⁡2A−2sin⁡Acos⁡Asin⁡2B]D=4\left[\frac{\sin ^{2} A+\cos ^{2} A-2 \sin A \cos A}{\sin ^{2} B}\right] D=4[(sin⁡A−cos⁡A)2sin⁡2B]D=4\left[\frac{(\sin A-\cos A)^{2}}{\sin ^{2} B}\right] D=[2(sin⁡A−cos⁡A)sin⁡B]2≥0D=\left[\frac{2(\sin A-\cos A)}{\sin B}\right]^{2} \geq 0 Therefore, roots of given equation are real.